Math, asked by allu2327, 5 months ago

do it with example.....​

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Answered by BrainlyEmpire
7

Matrix Multiplication

  • We are given a question involving Matrix Multiplication.

Matrix Product:-

  • Suppose A is a matrix of order \sf n\times m and B is a matrix of order \sf m\times p.
  • Let C be the product. That is, C = AB.

  • Then C is a matrix of order \sf n\times p where each element is given by:

  • \Large \text{$\displaystyle\sf c_{ij}=\sum\limits_{k=1}^{m} a_{ik}b_{kj}$}

Now let us look at the question. Consider the LHS:

\mathbb{LHS}\\\\\\ =\left[\begin{array}{ccc}\cos^{2}\theta&\cos\theta\sin\theta\\\cos\theta\sin\theta&\sin^{2}\theta\end{array}\right] \left[\begin{array}{ccc}\cos^{2}\phi&\cos\phi \sin\phi\\\cos\phi \sin\phi&\sin^{2}\phi\end{array}\right]

=\left[\begin{array}{ccc}\cos^2\theta\cos^2\phi+\cos\theta\sin\theta\cos\phi\sin\phi&\cos^2\theta\cos\phi\sin\phi+\cos\theta\sin\theta\sin^2\phi\\\cos\theta\sin\theta\cos^2\phi+\sin^2\theta\cos\phi\sin\phi&\cos\theta\sin\theta\cos\phi\sin\phi+\sin^2\theta\sin^2\phi\end{array}\right]

=\left[\begin{array}{ccc}\cos\theta\cos\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)&\cos\theta\sin\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)\\\sin\theta\cos\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)&\sin\theta\sin\phi(\cos\theta\cos\phi+\sin\theta\sin\phi)\end{array}\right]

=(\cos\theta\cos\phi+\sin\theta\sin\phi)\left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right]\\\\\\ \textsf{Use the identity: } \cos A\cos B+\sin A\sin B=\cos(A-B) \\\\\\ = \cos (\theta-\phi) \left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right]\\\\\\ \textsf{We are given that } \theta-\phi=\dfrac{\pi}{2}

=\cos\dfrac{\pi}{2} \left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right] \\\\\\ = 0 \times \left[\begin{array}{ccc}\cos\theta\cos\phi&\cos\theta\sin\phi\\\sin\theta\cos\phi&\sin\theta\sin\phi\end{array}\right] \\\\\\ = 0 \\\\\\ = \mathbb{RHS}

\mathcal{HENCE\ \ PROVED}

Answered by Anonymous
12

Matrix Product:-

Math

Suppose A is a matrix of order \sf n\times m and B is a matrix of order \sf m\times p .

Suppose A is a matrix of order

50 points

and B is a matrix of order

\sf n\times m

\sf m\times p

.

Let C be the product. That is, C = AB.

Let C be the product. That is, C = AB.

Then C is a matrix of order \sf n\times p where each element is given by:

Then C is a matrix of order

\sf n\times p

where each element is given by:

\Large \text{$\displaystyle\sf c_{ij}=\sum\limits_{k=1}^{m} a_{ik}b_{kj}$}

\Large \text{$\displaystyle\sf c_{ij}=\sum\limits_{k=1}^{m} a_{ik}b_{kj}$}

Now let us look at the question. Consider the LHS:

$$\begin{lgathered}\mathbb{LHS}\\\\\\ =\left[\begin{array}{ccc}\cos^{2}\theta&\cos\theta\sin\theta\\\cos\theta\sin\theta&\sin^{2}\theta\end{array}\right] \left[\begin{array}{ccc}\cos^{2}\phi&\cos\phi \sin\phi\\\cos\phi \sin\phi&\sin^{2}\phi\end{array}\right]\end{lgathered}$$

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